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;
template <typename T>
inline void ckmax(T& x, T y) {
  x = (y > x ? y : x);
}
template <typename T>
inline void ckmin(T& x, T y) {
  x = (y < x ? y : x);
}
const int MOD = 1e9 + 7;
inline int mod1(int x) { return x < MOD ? x : x - MOD; }
inline int mod2(int x) { return x < 0 ? x + MOD : x; }
inline void add(int& x, int y) { x = mod1(x + y); }
inline void sub(int& x, int y) { x = mod2(x - y); }
const int MAXN = 1e6;
int n, dp[MAXN + 5], dist[MAXN + 5], pos[MAXN + 5], nxt_1[MAXN + 5],
    nxt_d[MAXN + 5];
char s[MAXN + 5];
int main() {
  cin >> (s + 1);
  n = strlen(s + 1);
  int start_pos = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '1') {
      start_pos = i;
      break;
    }
  }
  if (!start_pos) {
    cout << n << endl;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '0') {
      dist[i] = dist[i - 1] + 1;
    } else {
      dist[i] = 0;
    }
  }
  for (int i = 0; i <= n + 1; ++i) pos[i] = n + 1;
  for (int i = n; i >= 1; --i) {
    nxt_1[i] = pos[0];
    nxt_d[i] = pos[dist[i] + 1];
    pos[dist[i]] = i;
  }
  dp[start_pos] = start_pos;
  for (int i = start_pos; i < n; ++i) {
    if (nxt_1[i] != n + 1) {
      add(dp[nxt_1[i]], dp[i]);
    }
    if (nxt_d[i] != n + 1) {
      add(dp[nxt_d[i]], dp[i]);
    }
  }
  int ans = 0;
  for (int i = start_pos; i <= n; ++i) {
    if (dist[i] <= dist[n]) {
      add(ans, dp[i]);
    }
  }
  cout << ans << endl;
  return 0;
}
