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;
string s;
const long long mod = 1e9 + 7;
long long dp[1000009];
long long seg[4000009];
void upd(long long v, long long tl, long long tr, long long idx,
         long long val) {
  if (tl == tr) {
    seg[v] = val;
    return;
  }
  if (idx <= (tl + tr) / 2)
    upd(2 * v, tl, (tl + tr) / 2, idx, val);
  else
    upd(2 * v + 1, (tl + tr) / 2 + 1, tr, idx, val);
  seg[v] = max(seg[2 * v], seg[2 * v + 1]);
}
long long get(long long v, long long tl, long long tr, long long val) {
  if (seg[v] < val) return -1;
  if (tl == tr) return tl;
  if (seg[2 * v + 1] >= val) return get(2 * v + 1, (tl + tr) / 2 + 1, tr, val);
  return get(2 * v, tl, (tl + tr) / 2, val);
}
vector<int> v;
long long bit[1000009];
long long dpsum;
void updbit(long long idx, long long val) {
  dpsum = (dpsum + val) % mod;
  for (++idx; idx <= v.size(); idx += idx & (-idx))
    bit[idx] = (bit[idx] + val) % mod;
}
long long getbit(long long idx) {
  long long ret = 0;
  for (++idx; idx; idx -= idx & (-idx)) ret = (ret + bit[idx]) % mod;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> s;
  int last = -1;
  long long mult = 1;
  for (int i = 0; i < s.size(); ++i)
    if (s[i] == '1') {
      if (last == -1) {
        mult = i - last;
        last = i;
        continue;
      }
      v.push_back(i - last);
      last = i;
    }
  mult = mult * (s.size() - last) % mod;
  if (v.empty()) {
    if (last == -1)
      cout << mult - 1 << '\n';
    else
      cout << mult << '\n';
    return 0;
  }
  for (long long i = 0; i < v.size(); ++i) {
    for (long long j = v[i]; j > 0; --j) {
      long long idx = get(1, 0, v.size() - 1, j);
      if (idx == -1) {
        dp[i] = (dp[i] + dpsum + 1) % mod;
      } else {
        dp[i] = (dp[i] + dpsum - getbit(idx - 1)) % mod;
        dp[i] = (dp[i] + mod) % mod;
      }
    }
    updbit(i, dp[i]);
    upd(1, 0, v.size() - 1, i, v[i]);
  }
  cout << mult * (dpsum + 1) % mod << '\n';
}