Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10, maxlg = 20;
const long long mod = 1e9 + 7;
string s;
int t0, a[maxn], n;
int nxt[maxn];
long long dp[maxn], part[maxn], ans;
void get_a() {
  cin >> s;
  s += '#';
  n = s.size() - 1;
  int po_a = 0;
  int po_s = 0;
  for (int i = 0; i <= n; i++) {
    if (i == n) {
      cout << n;
      exit(0);
    }
    if (s[i] == '1') {
      po_s = i;
      t0 = i;
      for (int j = i + 1; j <= n; j++)
        if (s[j] != '0') a[++po_a] = j - po_s - 1, po_s = j;
      break;
    }
  }
  n = po_a;
}
void get_nxt() {
  a[0] = 1e9;
  vector<int> q;
  q.push_back(0);
  for (int i = 1; i <= n; i++) {
    while (a[q.back()] <= a[i]) q.pop_back();
    nxt[i] = q.back();
    q.push_back(i);
  }
}
void get_dp() {
  part[0] = dp[0] = 1;
  for (int i = 1; i < n; i++) {
    int po = i - 1;
    for (int val = 0; val <= a[i]; val++) {
      while (a[nxt[po]] < val) po = nxt[po];
      if (val <= a[i - 1]) {
        dp[i] += dp[i - 1];
        continue;
      }
      dp[i] += part[i - 1];
      if (nxt[po] - 1 >= 0) dp[i] -= part[nxt[po] - 1];
    }
    dp[i] %= mod;
    if (dp[i] < 0) dp[i] += mod;
    part[i] = (part[i - 1] + dp[i]) % mod;
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  get_a();
  get_nxt();
  get_dp();
  ans = (part[n - 1] * (a[n] + 1) % mod * (t0 + 1)) % mod;
  cout << ans;
  return 0;
}