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>
namespace IO {
char gc() { return getchar(); }
template <typename Tp>
bool get1(Tp &x) {
  bool neg = 0;
  char c = gc();
  while (c != EOF && (c < '0' || c > '9') && c != '-') c = gc();
  if (c == '-') c = gc(), neg = 1;
  if (c == EOF) return false;
  x = 0;
  for (; c >= '0' && c <= '9'; c = gc()) x = x * 10 + c - '0';
  if (neg) x = -x;
  return true;
}
template <typename Tp>
void printendl(Tp x) {
  if (x < 0) putchar('-'), x = -x;
  static short a[40], sz;
  sz = 0;
  while (x > 0) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
  puts("");
}
}  // namespace IO
using IO::get1;
using IO::printendl;
const int inf = 0x3f3f3f3f;
const long long Linf = 1ll << 61;
const int mod = 1e9 + 7;
const int maxn = 1000111;
int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return ret;
}
int n, dp[maxn], pre[maxn];
char s[maxn];
int stk[maxn], top, sum[maxn], ways[maxn];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int s_ones = 0, last = 0, num = 0;
  int good = 1;
  while (s[good] == '0' && good <= n) good++;
  if (good > n) {
    printendl(n);
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (s[i] == s[i - 1])
      pre[i] = pre[i - 1];
    else
      pre[i] = i;
    if (s[i] == '0' && s[i + 1] != '0') {
      int len = i - pre[i] + 1;
      while (top && stk[top] <= len) top--;
      stk[++top] = len;
      sum[top] = sum[top - 1];
      ways[top] = s_ones;
      if (top > 1)
        sum[top] =
            (sum[top - 1] + 1ll * (stk[top - 1] - stk[top]) * ways[top - 1]) %
            mod;
      dp[i] = (sum[top] + 1ll * stk[top] * ways[top] + good - 1) % mod;
    } else if (s[i] == '1')
      s_ones = (s_ones + dp[pre[i] - 1] + 1) % mod;
  }
  int ans = (s_ones + (s[n] == '0' ? 1ll * ways[top] * stk[top] : 0)) % mod;
  printendl(ans);
  return 0;
}
