#include <bits/stdc++.h>
using namespace std;
using namespace std;
const long long mod2 = 998244353;
long long spf[100005];
long long fpow(long long x, long long y) {
  x = x % 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= 1000000007;
    y = y >> 1;
    x = x * x;
    x %= 1000000007;
  }
  return res;
}
long long inv(long long a, long long m = 1000000007) {
  long long c = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += c;
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  string s;
  cin >> s;
  long long a[n][2];
  for (int i = 0; i < n; i++) cin >> a[i][0] >> a[i][1];
  long long res = 0;
  for (int i = 0; i <= 100000; i++) {
    long long cnt = 0;
    for (int j = 0; j < n; j++) {
      long long x = s[j] - '0';
      if (i < a[j][1]) {
        if (x) cnt++;
      } else {
        long long y = i - a[j][1];
        y /= a[j][0];
        if (y % 2 == 0) {
          x ^= 1;
        }
        if (x) cnt++;
      }
    }
    res = max(res, cnt);
  }
  cout << res;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
using namespace std;
using namespace std;
const long long mod2 = 998244353;
long long spf[100005];
long long fpow(long long x, long long y) {
  x = x % 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= 1000000007;
    y = y >> 1;
    x = x * x;
    x %= 1000000007;
  }
  return res;
}
long long inv(long long a, long long m = 1000000007) {
  long long c = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += c;
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  string s;
  cin >> s;
  long long a[n][2];
  for (int i = 0; i < n; i++) cin >> a[i][0] >> a[i][1];
  long long res = 0;
  for (int i = 0; i <= 100000; i++) {
    long long cnt = 0;
    for (int j = 0; j < n; j++) {
      long long x = s[j] - '0';
      if (i < a[j][1]) {
        if (x) cnt++;
      } else {
        long long y = i - a[j][1];
        y /= a[j][0];
        if (y % 2 == 0) {
          x ^= 1;
        }
        if (x) cnt++;
      }
    }
    res = max(res, cnt);
  }
  cout << res;
}

```