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 M = (int)1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= M) {
    return a - M;
  }
  if (a < 0) {
    return a + M;
  }
  return a;
}
int mul(int a, int b) { return a * (long long)b % M; }
const int N = (int)1e6 + 7;
int n, v[N], cur, dp[N], pre[N];
string s;
map<int, int> last;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  for (auto &ch : s) {
    if (ch == '0') {
      cur++;
    } else {
      v[++n] = cur;
      cur = 0;
    }
  }
  if (n == 0) {
    cout << cur << "\n";
    return 0;
  }
  dp[1] = v[1] + 1;
  pre[1] = dp[1];
  last[v[1]] = last[M] = 1;
  for (int i = 2; i <= n; i++) {
    dp[i] = mul(pre[i - 1], v[i] + 1);
    for (int x = 0; x <= v[i]; x++) {
      dp[i] = add(dp[i], -pre[last.lower_bound(x)->second - 1]);
    }
    while (!last.empty() && last.begin()->first <= v[i]) {
      last.erase(last.begin());
    }
    last[v[i]] = i;
    pre[i] = add(pre[i - 1], dp[i]);
  }
  cout << mul(pre[n], cur + 1) << "\n";
  return 0;
}
