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
#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 3005;
const int mod = 1e9 + 7;

int n, m, a[MAXN], s[MAXN], f[MAXN][MAXN];
char c[MAXN];

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = Read(), m = Read();
	scanf("%s", c + 1);
	for (int i = 1; i <= n; i ++)
		a[i] = i, s[i] = s[i - 1] + (c[i] == '1');
	f[0][0] = 1;
	for (int i = 1, l, r; i <= m; i ++)
		l = Read(), r = Read(), a[l] = max(a[l], r);
	for (int i = 0, r = a[1]; i < n; i ++, r = max(r, a[i + 1]))
		for (int j = 0; j <= s[r]; j ++)
			if (f[i][j])
			{
				if (s[r] - j < r - i)
					f[i + 1][j] = (f[i + 1][j] + f[i][j]) % mod;
				if (s[r] - j)
					f[i + 1][j + 1] = (f[i + 1][j + 1] + f[i][j]) % mod;
			}
	return printf("%d\n", f[n][s[n]]), 0;
}
