You are given a program that consists of n instructions. Initially a single variable x is assigned to 0. Afterwards, the instructions are of two types: 

  * increase x by 1; 
  * decrease x by 1. 



You are given m queries of the following format: 

  * query l r — how many distinct values is x assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order? 

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

Then the description of t testcases follows.

The first line of each testcase contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of instructions in the program and the number of queries.

The second line of each testcase contains a program — a string of n characters: each character is either '+' or '-' — increment and decrement instruction, respectively.

Each of the next m lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the description of the query.

The sum of n over all testcases doesn't exceed 2 ⋅ 10^5. The sum of m over all testcases doesn't exceed 2 ⋅ 10^5. 

Output

For each testcase print m integers — for each query l, r print the number of distinct values variable x is assigned to if all the instructions between the l-th one and the r-th one inclusive are ignored and the rest are executed without changing the order.

Example

Input


2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4


Output


1
2
4
4
3
3
4
2
3
2
1
2
2
2

Note

The instructions that remain for each query of the first testcase are: 

  1. empty program — x was only equal to 0; 
  2. "-" — x had values 0 and -1; 
  3. "---+" — x had values 0, -1, -2, -3, -2 — there are 4 distinct values among them; 
  4. "+--+--+" — the distinct values are 1, 0, -1, -2. 
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,Q;
int l[200005],r[200005],mnl[200005],mnr[200005],mxl[200005],mxr[200005];
char s[200005];
int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	int T; cin >> T;
	while(T--) {
		cin >> n >> Q >> s+1;
		for(int i = 0;i <= n+1;i++) l[i] = r[i] = 0, mnl[i] = mnr[i] = 0, mxl[i] = mxr[i] = 0;
		for(int i = 1;i <= n;i++) l[i] = l[i-1]+(s[i] == '+' ? 1 : -1);
		for(int i = n;i >= 1;i--) r[i] = r[i+1]+(s[i] == '-' ? 1 : -1);
		for(int i = 1;i <= n;i++) {
			mnl[i] = min(mnl[i-1],l[i]);
			mxl[i] = max(mxl[i-1],l[i]);
		}
		for(int i = n;i >= 1;i--) {
			mnr[i] = min(mnr[i+1],r[i]);
			mxr[i] = max(mxr[i+1],r[i]);
		}
		while(Q--) {
			int L,R; cin >> L >> R;
			int now = l[L-1];
			int mn = mnl[L-1], mx = mxl[L-1];
			mn = min(mn,mnr[R+1]+now-r[R+1]), mx = max(mx,mxr[R+1]+now-r[R+1]);
			cout << mx-mn+1 << '\n';
		}
	}
}