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>
template <class T>
inline void read(T &x) {
  static char ch;
  static bool opt;
  while (!isdigit(ch = getchar()) && ch != '-')
    ;
  x = (opt = ch == '-') ? 0 : ch - '0';
  while (isdigit(ch = getchar())) x = x * 10 + ch - '0';
  if (opt) x = ~x + 1;
}
template <class T>
inline void putint(T x) {
  static char buf[45], *tail = buf;
  if (!x)
    putchar('0');
  else {
    if (x < 0) putchar('-'), x = ~x + 1;
    for (; x; x /= 10) *++tail = x % 10 + '0';
    for (; tail != buf; --tail) putchar(*tail);
  }
}
template <class T>
inline bool tense(T &x, const T &y) {
  return y < x ? x = y, true : false;
}
template <class T>
inline bool relax(T &x, const T &y) {
  return x < y ? x = y, true : false;
}
template <class T>
inline T getAbs(const T &x) {
  return x < 0 ? -x : x;
}
const int mod = 1e9 + 7;
inline void add(int &x, const int &y) {
  if (x += y, x >= mod) x -= mod;
}
inline void dec(int &x, const int &y) {
  if (x -= y, x < 0) x += mod;
}
const int MaxN = 1e6 + 5;
int n;
int f[MaxN];
char s[MaxN];
int pre[MaxN];
int nxt[MaxN][2];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  bool all0 = true;
  for (int i = n - 1; i >= 0; --i) {
    nxt[i][0] = nxt[i + 1][0];
    nxt[i][1] = nxt[i + 1][1];
    nxt[i][s[i + 1] - '0'] = i + 1;
  }
  for (int i = 1; i <= n; ++i) {
    pre[i] = s[i] == '0' ? pre[i - 1] + 1 : 0;
  }
  for (int i = 1; i <= n; ++i) {
    all0 &= s[i] == '0';
  }
  if (all0) {
    printf("%d\n", n);
    return 0;
  }
  static int stk[MaxN], top;
  for (int i = n, sum = 0; i >= 1; --i) {
    if (s[i] == '0') {
      f[i] = pre[i] < i && pre[i] <= pre[n];
      if (nxt[i][1]) {
        add(f[i], f[nxt[i][1]]);
      }
      while (top && pre[stk[top]] <= pre[i]) dec(sum, f[stk[top--]]);
      add(sum, f[stk[++top] = i]);
    } else {
      add(f[i] = (bool)i, sum);
      if (nxt[i][1]) {
        add(f[i], f[nxt[i][1]]);
      }
    }
  }
  int res = 0;
  for (int i = 1; i <= n; ++i) {
    add(res, f[i]);
    if (s[i] == '1') break;
  }
  std::cout << res << '\n';
  return 0;
}