Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, ..., N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded. Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.

　　The first line contains an integer T, indicating the number of test cases.
　　For each test case, the first line contains two integers N(1 < N < 50001) and M(1 < M < 50001). N is the number of vases, and M is the operations of Alice. Each of the next M lines contains three integers. The first integer of one line is K(1 or 2). If K is 1, then two integers A and F follow. It means Alice receive F flowers and try to put a flower in the vase A first. If K is 2, then two integers A and B follow. It means the owner would like to clean the vases numbered from A to B(A <= B).

　　For each operation of which K is 1, output the position of the vase in which Alice put the first flower and last one, separated by a blank. If she can not put any one, then output 'Can not put any one.'. For each operation of which K is 2, output the number of discarded flowers. 
　　Output one blank line after each test case.

Sample Output

Can not put any one.

 3

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 #include <string>

 #include <math.h>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <set>

 #include <stack>

 #include <map>

 #include <math.h>

 const int INF=0x3f3f3f3f;

 typedef long long LL;

 const int mod=1e9+;

 const int maxn=1e5+;

 using namespace std;

 struct SegTree_node

 {

     int l;

     int r;

     int num;//区间内空瓶的数量

     int tag;//如果为-1则为初始状态,不用向下更新，如果为0意思将子区间内的花清空，为1为插满花

 }SegTree[<<]; 

 int n,m;

 void PushUp(int rt)//向上传递区间信息

 {

     SegTree[rt].num=SegTree[rt<<].num+SegTree[rt<<|].num;

 }

 void PushDown(int rt)//向下传递状态

 {

     if(SegTree[rt].tag!=-)

     {

         SegTree[rt<<].tag=SegTree[rt<<|].tag=SegTree[rt].tag;

         int mid=(SegTree[rt].l+SegTree[rt].r)>>;

         SegTree[rt<<].num=(mid-SegTree[rt].l+)*SegTree[rt].tag;

         SegTree[rt<<|].num=(SegTree[rt].r-mid)*SegTree[rt].tag;

         SegTree[rt].tag=-;

     }

 }

 void Build(int l,int r,int rt)

 {

     SegTree[rt].l=l;

     SegTree[rt].r=r;

     SegTree[rt].num=r-l+;

     SegTree[rt].tag=-;

     if(l==r)

         return ;

     int mid=(l+r)>>;

     Build(l,mid,rt<<);

     Build(mid+,r,rt<<|);

 }

 void Update(int L,int R,int f,int rt)

 {

     int l=SegTree[rt].l;

     int r=SegTree[rt].r;

     if(L<=l&&R>=r)

     {

         SegTree[rt].num=(r-l+)*f;//先计算

         SegTree[rt].tag=f;//标记当前区间

         return ;

     }

     if(l==r)//不要忘记

         return ;

     PushDown(rt);//向下传递状态

     int mid=(l+r)>>;

     if(L<=mid)

         Update(L,R,f,rt<<);

     if(R>mid)

         Update(L,R,f,rt<<|);

     PushUp(rt);

 }

 int Query(int L,int R,int rt)//得到区间[L,R]中的空花瓶数

 {

     int l=SegTree[rt].l;

     int r=SegTree[rt].r;

     if(R<l||L>r)

         return ;

     if(L<=l&&R>=r)

         return SegTree[rt].num;

     PushDown(rt);

     int mid=(l+r)>>;

     int sum=;

     if(L<=mid)

         sum+=Query(L,R,rt<<);

     if(R>mid)

         sum+=Query(L,R,rt<<|);

     return sum;

 }

 int dive(int x,int num)//二分查找，第一个为开始的位置，第二个参数为要插花的个数

 {

     int l=x;

     int r=n;

     int ans=;

     while(l<=r)

     {

         int mid=(l+r)>>;

         if(Query(x,mid,)>=num)

         {

             ans=mid;

             r=mid-;

         }

         else

             l=mid+;

     }

     return ans;

 }

 int main()

 {

     int T;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d %d",&n,&m);

         Build(,n,);

         for(int i=;i<=m;i++)

         {

             int op;

             scanf("%d",&op);

             if(op==)

             {

                 int A,F;

                 scanf("%d %d",&A,&F);

                 A++;//存储时序号从一开始，故序号要全加一，下同

                 int cnt=Query(A,n,);//得到区间[A, n]中的空花瓶数

                 if(cnt==)

                     printf("Can not put any one.\n");

                 else

                 {

                     int L=dive(A,);//二分左端点(第1个能插花的位置)

                     int R=dive(A,min(cnt,F));//二分右端点(最后一个能插花的位置)

                     Update(L,R,,);//将区间[L,R]的花瓶全部插满

                     printf("%d %d\n",L-,R-);

                 }

             }

             else if(op==)

             {

                 int A,B;

                 scanf("%d %d",&A,&B);

                 A++;

                 B++;

                 printf("%d\n",B-A+-Query(A,B,));

                 Update(A,B,,);//将区间[A,B]的花瓶全部清空

             }

         }

         printf("\n");

     }

     return ;

 }