Problem: You are given a 4x4 grid. You play a game — there is a sequence of tiles, each of them is either 2x1 or 1x2. Your task is to consequently place all tiles from the given sequence in the grid. When tile is placed, each cell which is located in fully occupied row or column is deleted (cells are deleted at the same time independently). You can place tile in the grid at any position, the only condition is that tiles (and tile parts) should not overlap. Your goal is to proceed all given figures and avoid crossing at any time.

Input

The only line contains a string s consisting of zeroes and ones (1 ≤ |s| ≤ 1000). Zero describes vertical tile, one describes horizontal tile.

Output

Output |s| lines — for each tile you should output two positive integers r,c, not exceeding 4, representing numbers of smallest row and column intersecting with it.

If there exist multiple solutions, print any of them.

Example

Input


010


Output


1 1
1 2
1 4

Note

Following image illustrates the example after placing all three tiles: 

<image> Then the first row is deleted:  <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long int modadd(long long int n, long long int m,
                            long long int p = 1000000007) {
  return ((n + m) % p + p) % p;
}
inline long long int modsub(long long int n, long long int m,
                            long long int p = 1000000007) {
  return ((n - m + p) % p + p) % p;
}
inline long long int modpro(long long int n, long long int m,
                            long long int p = 1000000007) {
  return (((n % p) * (m % p)) % p + p) % p;
}
unsigned long long int powe(long long int first, long long int second) {
  unsigned long long int res = 1;
  while (second > 0) {
    if (second & 1) res = res * first;
    second = second >> 1;
    first = first * first;
  }
  return res;
}
long long int modpow(long long int first, long long int second,
                     long long int p = 1000000007) {
  long long int res = 1;
  while (second > 0) {
    if (second & 1) res = modpro(res, first, p);
    second = second >> 1;
    first = modpro(first, first, p);
  }
  return res;
}
inline long long int modInverse(long long int n, long long int p = 1000000007) {
  if (n == 1) return 1;
  return modpow(n, p - 2, p);
}
inline long long int moddiv(long long int n, long long int m,
                            long long int p = 1000000007) {
  return modpro(n, modInverse(m, p), p);
}
inline long long int modadd3(long long int first, long long int second,
                             long long int z, long long int p = 1000000007) {
  return modadd(modadd(first, second, p), z, p);
}
inline long long int modadd4(long long int first, long long int second,
                             long long int z, long long int w,
                             long long int p = 1000000007) {
  return modadd(modadd(first, second, p), modadd(z, w, p), p);
}
inline long long int modnCr(long long int fac[], int n, int r,
                            long long int p = 1000000007) {
  if (r == 0) return 1;
  return modpro(fac[n], modInverse(modpro(fac[r], fac[n - r], p), p), p);
}
template <typename T>
inline T max3(T first, T second, T z) {
  return max(max(first, second), z);
}
template <typename T>
inline T max4(T first, T second, T z, T w) {
  return max(max3(first, second, w), z);
}
template <typename T>
inline T min3(T first, T second, T z) {
  return min(min(first, second), z);
}
template <typename T>
inline T min4(T first, T second, T z, T w) {
  return min(min3(first, second, w), z);
}
template <typename T>
void printArr(T *arr, int s, int n) {
  for (int i = s; i <= n; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
pair<int, int> ans[2005];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int erer = 1;
  for (int erer2 = (1); erer2 < (erer + 1); erer2++) {
    string s;
    cin >> s;
    int f = 0, g = 0;
    for (int i = (0); i < (((long long int)(s).size())); i++) {
      if (s[i] == '0') {
        if (f) {
          ans[i] = {3, 1};
        } else {
          ans[i] = {1, 1};
        }
        f ^= 1;
      }
      if (s[i] == '1') {
        ans[i] = {g + 1, 3};
        g++;
        g %= 4;
      }
    }
    for (int i = (0); i < (((long long int)(s).size())); i++)
      cout << (ans[i].first) << " " << (ans[i].second) << endl;
  }
  return 0;
}