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;
string s;
long long dp[1000005][2];
long long succ[2][1000005];
long long run[1000005];
long long rpos[1000005];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  long long n = s.size();
  s = "#" + s;
  bool flag = false;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') flag = true;
  }
  if (!flag) {
    cout << n << endl;
    return 0;
  }
  long long cnt = 0;
  for (int i = 0; i <= n; i++) {
    if (s[i] == '0')
      cnt++;
    else
      cnt = 0;
    run[i] = cnt;
  }
  long long nx = n + 1;
  for (int i = 1; i <= n + 2; i++) rpos[i] = n + 1;
  for (int i = n; i >= 0; i--) {
    succ[0][i] = rpos[run[i] + 1];
    succ[1][i] = nx;
    if (s[i] == '1')
      nx = i;
    else
      rpos[run[i]] = i;
  }
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 2; j++) {
      if (succ[0][i] <= n) {
        if (j == 1 || s[i + 1] == '0')
          (dp[succ[0][i]][j] += dp[i][j]) %= 1000000007;
      }
      if (succ[1][i] <= n) (dp[succ[1][i]][1] += dp[i][j]) %= 1000000007;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '1') ans += dp[i][1], ans %= 1000000007;
    if (s[i] == '0' && run[i] <= run[n]) ans += dp[i][1], ans %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}
