#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
inline int nxt() {
  int aaa;
  scanf("%d", &aaa);
  return aaa;
}
inline long long lxt() {
  long long aaa;
  scanf("%lld", &aaa);
  return aaa;
}
inline double dxt() {
  double aaa;
  scanf("%lf", &aaa);
  return aaa;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
}
int cnt[105][10005];
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1')
      cnt[i][0] = 1;
    else
      cnt[i][0] = 2;
  }
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  for (int i = 0; i < n; i++) {
    cnt[i][v[i].second] = (cnt[i][0] == 1 ? 2 : 1);
    for (int k = v[i].first + v[i].second; k <= 10000; k += v[i].first) {
      cnt[i][k] = (cnt[i][k - v[i].first] == 1 ? 2 : 1);
    }
    for (int k = 1; k <= 10000; k++) {
      if (cnt[i][k] == 0) {
        cnt[i][k] = cnt[i][k - 1];
      }
    }
  }
  int ans = 0;
  for (int i = 0; i <= 10000; i++) {
    int subans = 0;
    for (int j = 0; j < n; j++) {
      if (cnt[j][i] == 1) subans++;
    }
    ans = max(subans, ans);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(10);
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}

### 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;
int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
inline int nxt() {
  int aaa;
  scanf("%d", &aaa);
  return aaa;
}
inline long long lxt() {
  long long aaa;
  scanf("%lld", &aaa);
  return aaa;
}
inline double dxt() {
  double aaa;
  scanf("%lf", &aaa);
  return aaa;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
}
int cnt[105][10005];
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1')
      cnt[i][0] = 1;
    else
      cnt[i][0] = 2;
  }
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  for (int i = 0; i < n; i++) {
    cnt[i][v[i].second] = (cnt[i][0] == 1 ? 2 : 1);
    for (int k = v[i].first + v[i].second; k <= 10000; k += v[i].first) {
      cnt[i][k] = (cnt[i][k - v[i].first] == 1 ? 2 : 1);
    }
    for (int k = 1; k <= 10000; k++) {
      if (cnt[i][k] == 0) {
        cnt[i][k] = cnt[i][k - 1];
      }
    }
  }
  int ans = 0;
  for (int i = 0; i <= 10000; i++) {
    int subans = 0;
    for (int j = 0; j < n; j++) {
      if (cnt[j][i] == 1) subans++;
    }
    ans = max(subans, ans);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.setf(ios::fixed);
  cout.precision(10);
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}

```