Problem: There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e3 + 10;
template <class T>
inline void _read(T &x)
{
	x = 0;
	char t = getchar();
	while (!isdigit(t)) t = getchar();
	while (isdigit(t))
	{
		x = x * 10 + t - '0';
		t = getchar();
	}
}
int n, m;
char ch[MAXN];
int sum[MAXN], f[MAXN][MAXN], las[MAXN], tot;
const int mo = 1e9 + 7;
int add(int x, int y)
{
	return x + y >= mo ? x + y - mo : x + y;
}
int main()
{
	_read(n), _read(m);
	scanf("%s", ch + 1);
	for (int i = 1; i <= n + 1; ++i)
	{
		sum[i] = sum[i - 1], las[i] = i;
		if (ch[i] == '1') ++tot, ++sum[i];
	}
	for (int i = 1, x, y; i <= m; ++i) _read(x), _read(y), las[x] = max(las[x], y);
	for(int i = 1; i <= n + 1; ++i)las[i] = max(las[i], las[i - 1]);
	f[1][sum[las[1]]] = 1;
	for(int i = 1; i <= n; ++i)
		for(int j = 0; j <= tot; ++j)
			if(f[i][j])
			{
				int a = las[i], b = las[i + 1];
				if(j)f[i + 1][j + sum[b] - sum[a] - 1] = add(f[i + 1][j + sum[b] - sum[a] - 1], f[i][j]);
				if(las[i] + 1 - i - j) f[i + 1][j + sum[b] - sum[a]] = (f[i + 1][j + sum[b] - sum[a]], f[i][j]);
			}
	printf("%d\n", f[n + 1][0]);
}