Problem: This task is very simple. Given a string S of length n and q queries each query is on the format i j k which means sort the substring consisting of the characters from i to j in non-decreasing order if k = 1 or in non-increasing order if k = 0.

Output the final string after applying the queries.

Input

The first line will contain two integers n, q (1 ≤ n ≤ 105, 0 ≤ q ≤ 50 000), the length of the string and the number of queries respectively. 

Next line contains a string S itself. It contains only lowercase English letters.

Next q lines will contain three integers each i, j, k (1 ≤ i ≤ j ≤ n, <image>).

Output

Output one line, the string S after applying the queries.

Examples

Input

10 5
abacdabcda
7 10 0
5 8 1
1 4 0
3 6 0
7 10 1


Output

cbcaaaabdd

Input

10 1
agjucbvdfk
1 10 1


Output

abcdfgjkuv

Note

First sample test explanation:

<image>

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<image>

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct ST {
  int sum[4 * 100000 + 2];
  int lazy[4 * 100000 + 2];
  ST() {
    for (int i = 0; i < 4 * 100000 + 2; i++) sum[i] = 0, lazy[i] = -1;
  }
  void propagate(int node, int l, int r) {
    if (lazy[node] == -1) return;
    for (int i = 0; i < 2; i++) {
      lazy[2 * node + i] = lazy[node];
      sum[2 * node + i] =
          lazy[node] * (((r - l + 1) >> 1) + ((r - l + 1) & 1) * (1 - i));
    }
    lazy[node] = -1;
    return;
  }
  void update(int node, int l, int r, int a, int b, int val) {
    if (l > b || r < a) return;
    if (l >= a && r <= b) {
      lazy[node] = val;
      sum[node] = (r - l + 1) * val;
      return;
    }
    propagate(node, l, r);
    int mid = (l + r) >> 1;
    update(2 * node + 0, l, mid + 0, a, b, val);
    update(2 * node + 1, mid + 1, r, a, b, val);
    sum[node] = sum[2 * node] + sum[2 * node + 1];
    return;
  }
  int query(int node, int l, int r, int a, int b) {
    if (l > b || r < a) return 0;
    if (l >= a && r <= b) return sum[node];
    propagate(node, l, r);
    int mid = (l + r) >> 1;
    return query(2 * node, l, mid, a, b) +
           query(2 * node + 1, mid + 1, r, a, b);
  }
};
int n, q;
char s[100000 + 2];
ST alfa[26];
int main() {
  scanf("%d %d\n", &n, &q);
  gets(s);
  for (int i = 0; i < n; i++) alfa[s[i] - 'a'].update(1, 1, n, i + 1, i + 1, 1);
  int l, r, k;
  while (q--) {
    scanf("%d %d %d", &l, &r, &k);
    if (k) {
      int now = l;
      for (int i = 0; i < 26; i++) {
        int tmp = alfa[i].query(1, 1, n, l, r);
        alfa[i].update(1, 1, n, l, r, 0);
        alfa[i].update(1, 1, n, now, now + tmp - 1, 1);
        now += tmp;
      }
    } else {
      int now = r;
      for (int i = 0; i < 26; i++) {
        int tmp = alfa[i].query(1, 1, n, l, r);
        alfa[i].update(1, 1, n, l, r, 0);
        alfa[i].update(1, 1, n, now - tmp + 1, now, 1);
        now -= tmp;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 26; j++)
      if (alfa[j].query(1, 1, n, i + 1, i + 1)) {
        printf("%c", j + 'a');
        break;
      }
  }
  printf("\n");
  return 0;
}