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 = 1000000007;
const int N = 1000000;
int n;
char a[N + 5];
vector<pair<int, bool> > v;
int dp[N + 1];
pair<pair<int, int>, int> stk[N];
int top;
int main() {
  scanf("%s", a + 1);
  n = strlen(a + 1);
  for (int i = 1, ie; i <= n; i = ie + 1) {
    ie = i;
    while (ie + 1 <= n && a[ie + 1] == a[i]) ie++;
    v.push_back(make_pair(ie - i + 1, a[i] ^ 48));
  }
  int sum = 0, now = 0, las = !v.back().second * v.back().first;
  dp[v.size()] = 1;
  for (int i = v.size() - 1; ~i; i--) {
    if (v[i].second == 0) {
      dp[i] = 1ll * v[i].first * dp[i + 1] % mod;
      while (top && stk[top - 1].first.second <= v[i].first)
        (now -= 1ll *
                (stk[top - 1].first.second - stk[top - 1].first.first + 1) *
                stk[top - 1].second % mod) %= mod,
            top--;
      if (top)
        (now -= 1ll * (v[i].first + 1 - stk[top - 1].first.first) *
                stk[top - 1].second % mod) %= mod,
            stk[top - 1].first.first = v[i].first + 1;
      stk[top++] =
          make_pair(make_pair(1, v[i].first), (i < v.size() - 1) * dp[i + 1]),
      (now += (i < v.size() - 1) * dp[i]) %= mod;
    } else
      dp[i] = (sum += 1ll * v[i].first * (now + 1 + las) % mod) %= mod;
  }
  if (v.size() == 1)
    cout << (dp[0] + mod) % mod;
  else if (v[0].second == 1)
    cout << (dp[0] + mod) % mod;
  else
    cout << ((dp[0] + dp[1]) % mod + mod) % mod;
  return 0;
}
