Koa the Koala has a binary string s of length n. Koa can perform no more than n-1 (possibly zero) operations of the following form:

In one operation Koa selects positions i and i+1 for some i with 1 ≤ i < |s| and sets s_i to max(s_i, s_{i+1}). Then Koa deletes position i+1 from s (after the removal, the remaining parts are concatenated).

Note that after every operation the length of s decreases by 1.

How many different binary strings can Koa obtain by doing no more than n-1 (possibly zero) operations modulo 10^9+7 (1000000007)?

Input

The only line of input contains binary string s (1 ≤ |s| ≤ 10^6). For all i (1 ≤ i ≤ |s|) s_i = 0 or s_i = 1.

Output

On a single line print the answer to the problem modulo 10^9+7 (1000000007).

Examples

Input


000


Output


3


Input


0101


Output


6


Input


0001111


Output


16


Input


00101100011100


Output


477

Note

In the first sample Koa can obtain binary strings: 0, 00 and 000.

In the second sample Koa can obtain binary strings: 1, 01, 11, 011, 101 and 0101. For example:

  * to obtain 01 from 0101 Koa can operate as follows: 0101 → 0(10)1 → 011 → 0(11) → 01. 
  * to obtain 11 from 0101 Koa can operate as follows: 0101 → (01)01 → 101 → 1(01) → 11. 



Parentheses denote the two positions Koa selected in each operation.
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int sum(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
int sub(int a, int b) { return a - b < 0 ? a - b + MOD : a - b; }
int mul(int a, int b) { return (1LL * a * b) % MOD; }
int pw(int a, int b) {
  if (!b) return 1;
  int r = pw(a, b / 2);
  r = mul(r, r);
  if (b % 2) r = mul(r, a);
  return r;
}
const int MAXN = 1e6 + 7;
int shuru[MAXN], shesh[MAXN];
int zero[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  zero[0] = 1;
  int zerosum = 1;
  int sheshsum = 0;
  shuru[n] = 0;
  int z = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '0') {
      z++;
      if (i + z == n) zerosum = sum(zerosum, 1);
      zerosum = sub(zerosum, zero[z]);
      zero[z] = shuru[i + z];
      zerosum = sum(zerosum, zero[z]);
    } else {
      z = 0;
      shesh[i] = zerosum;
      sheshsum = sum(sheshsum, shesh[i]);
      shuru[i] = sheshsum;
    }
  }
  if (z == n) {
    cout << n << "\n";
  } else {
    cout << mul(z + 1, shuru[z]) << "\n";
  }
  return 0;
}
