Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 100005;
const long long mod = 1e9 + 7;
long long sum[maxn << 2];
void build(long long rt, long long l, long long r) {
  if (l == r) {
    scanf("%1lld", &sum[rt]);
  } else {
    long long mid = (l + r) / 2;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  }
}
long long query(long long rt, long long l, long long r, long long L,
                long long R) {
  if (L <= l && r <= R) {
    return sum[rt];
  }
  long long mid = (l + r) / 2;
  long long ans = 0;
  if (L <= mid) ans += query(rt << 1, l, mid, L, R);
  if (mid < R) ans += query(rt << 1 | 1, mid + 1, r, L, R);
  return ans;
}
long long qpow(long long a, long long x) {
  long long ret = 1;
  while (x) {
    if (x & 1) ret = ret * a % mod;
    a = a * a % mod;
    x >>= 1;
  }
  return ret;
}
int main() {
  long long n, q;
  scanf("%lld%lld", &n, &q);
  build(1, 1, n);
  while (q--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    long long x = query(1, 1, n, l, r);
    long long y = (r - l + 1) - x;
    long long ans = (qpow(2, x) - 1 + mod) % mod * qpow(2, y) % mod;
    printf("%lld\n", ans);
  }
  return 0;
}