#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int mod = 1e9 + 7;
clock_t t1, t2;
long long binPow(long long a, long long b) {
  long long x = 1, y = a;
  while (b) {
    if (b & 1) x = (x * y) % mod;
    y = (y * y) % mod;
    b >>= 1;
  }
  return x % mod;
}
long long inverserEuler(int n) { return binPow(n, mod - 2); }
long long C(int k, int n) {
  vector<long long> f(n + 1, 1);
  for (int i = 2; i <= n; i++) f[i] = (f[i - 1] * i) % mod;
  return (f[n] * ((inverserEuler(f[k]) * inverserEuler(f[n - k])) % mod) %
          mod) %
         mod;
}
const int maxn = 1e7;
int cnt[maxn + 5];
void trunghieu() {
  int n;
  string s;
  cin >> n >> s;
  int i, j;
  vector<pair<int, int>> lights;
  int ans = 0;
  for (i = 0; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    lights.push_back({a, b});
  }
  for (j = 0; j < n; ++j) {
    int x = (s[j] == '1') ? 1 : 0;
    for (i = 0; i < 1e6; ++i) {
      if (i == lights[j].second) {
        x = 1 - x;
      } else if (i > lights[j].second) {
        if ((i - lights[j].second) % lights[j].first == 0) {
          x = 1 - x;
        }
      }
      cnt[i] += (x == 1);
      ans = max(ans, cnt[i]);
    }
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  trunghieu();
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.

After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.

You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.

<image> Here is a graphic for the first example.

Input

The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.

The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".

The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.

Output

Print a single integer — the maximum number of lights that will ever be on at the same time.

Examples

Input


3
101
3 3
3 2
3 1


Output


2


Input


4
1111
3 4
5 2
3 1
3 2


Output


4


Input


6
011100
5 3
5 5
2 4
3 5
4 2
1 5


Output


6

Note

For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).

In the second example, all lights are initially on. So the answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int mod = 1e9 + 7;
clock_t t1, t2;
long long binPow(long long a, long long b) {
  long long x = 1, y = a;
  while (b) {
    if (b & 1) x = (x * y) % mod;
    y = (y * y) % mod;
    b >>= 1;
  }
  return x % mod;
}
long long inverserEuler(int n) { return binPow(n, mod - 2); }
long long C(int k, int n) {
  vector<long long> f(n + 1, 1);
  for (int i = 2; i <= n; i++) f[i] = (f[i - 1] * i) % mod;
  return (f[n] * ((inverserEuler(f[k]) * inverserEuler(f[n - k])) % mod) %
          mod) %
         mod;
}
const int maxn = 1e7;
int cnt[maxn + 5];
void trunghieu() {
  int n;
  string s;
  cin >> n >> s;
  int i, j;
  vector<pair<int, int>> lights;
  int ans = 0;
  for (i = 0; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    lights.push_back({a, b});
  }
  for (j = 0; j < n; ++j) {
    int x = (s[j] == '1') ? 1 : 0;
    for (i = 0; i < 1e6; ++i) {
      if (i == lights[j].second) {
        x = 1 - x;
      } else if (i > lights[j].second) {
        if ((i - lights[j].second) % lights[j].first == 0) {
          x = 1 - x;
        }
      }
      cnt[i] += (x == 1);
      ans = max(ans, cnt[i]);
    }
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  trunghieu();
  return 0;
}

```