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;
void solve(int m, string k)
{
	vector<int>ma(k.size()), mi(k.size()), com(k.size());
	vector<int>maR(k.size() + 1), miR(k.size() + 1);
	com[0] = 0; ma[0] = 0; mi[0] = 0;
	for (int i = 1; i < k.size(); i++)
	{
		if (k[i - 1] == '+')
		{
			com[i] = com[i - 1] + 1;
			ma[i] = max(ma[i - 1], com[i]);
			mi[i] = min(mi[i - 1], com[i]);
		}
		else
		{
			com[i] = com[i - 1] - 1;
			ma[i] = max(ma[i - 1], com[i]);
			mi[i] = min(mi[i - 1], com[i]);
		}

	}
	if (k[k.size() - 1] == '+')
	{maR[k.size()] = com[k.size() - 1] + 1;
    }
	else
	{
		maR[k.size()] = com[k.size() - 1] - 1;
	}
		miR[k.size()] = maR[k.size()];
		for (int i = k.size() - 1; i >= 0; i--)
		{
			maR[i] = max(maR[i + 1], com[i]);
			miR[i] = min(miR[i + 1], com[i]);
		}
		for (int i = 0; i < m; i++)
		{
			int x, y;
			cin >> x >> y;
			if (x == 1 && y == k.size())
			{
				cout << "1" << endl;
			}
			else if (x==1)
			{
				int diff = 0 - com[y];
				cout << (maR[y] + diff) - (miR[y] + diff) + 1 << endl;
			}
			else if (y == k.size())
			{
				cout << (ma[x - 1] - mi[x - 1]) + 1 << endl;
			}
			else
			{
				int mam = ma[x -2];
				int mim = mi[x - 2];
				int mam2 = maR[y];
				int mim2 = miR[y];
				int diff = com[x - 1] - com[y];
				mam2 += diff;
				mim2 += diff;
				mam = max(mam, mam2);
				mim = min(mim, mim2);
				cout << mam - mim + 1 << endl;
			}
		}
}
int main()
{
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int t;
	cin >> t;
	while (t--)
	{
		int n, m;
		cin >> n >> m;
		string k;
		cin >> k;
		solve(m, k);
	}
}