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>

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
namespace thecold {
inline int read() {
  int res = 0;
  bool flag = false;
  char c = getchar();
  while (!isdigit(c)) flag = (c == '-'), c = getchar();
  while (isdigit(c)) res = res * 10 + (c ^ 48), c = getchar();
  return !flag ? res : -res;
}
const int Max_n = 1e5 + 5;
struct node {
  int l, r, sum, tag;
  node(int a, int b) { l = a, r = b, sum = 0, tag = -1; }
  node() {}
} tree[Max_n << 2];
struct Node {
  int l, r, type;
  Node() {}
  Node(const int a, const int b, const int c) { l = a, r = b, type = c; }
} opt[Max_n];
char c[Max_n];
int a[Max_n], n, m, ans[30][Max_n], lim;
inline void push_up(const int tr) {
  tree[tr].sum = tree[tr * 2].sum + tree[tr * 2 + 1].sum;
}
inline void push_down(const int tr) {
  if (tree[tr].tag == -1) return;
  tree[tr * 2].tag = tree[tr * 2 + 1].tag = tree[tr].tag;
  tree[tr * 2].sum = (tree[tr * 2].r - tree[tr * 2].l + 1) * tree[tr].tag;
  tree[tr * 2 + 1].sum =
      (tree[tr * 2 + 1].r - tree[tr * 2 + 1].l + 1) * tree[tr].tag;
  tree[tr].tag = -1;
}
void build(const int l, const int r, const int tr) {
  tree[tr] = node(l, r);
  if (l == r) return tree[tr].sum = (a[l] >= lim), void();
  int mid = (l + r) >> 1;
  build(l, mid, tr * 2), build(mid + 1, r, tr * 2 + 1);
  push_up(tr);
}
int query(const int l, const int r, const int tr) {
  int ll = tree[tr].l, rr = tree[tr].r;
  if (ll > r || rr < l) return 0;
  if (l <= ll && rr <= r) return tree[tr].sum;
  push_down(tr);
  return query(l, r, tr * 2) + query(l, r, tr * 2 + 1);
}
void change(const int l, const int r, const int tr, const int val) {
  int ll = tree[tr].l, rr = tree[tr].r;
  if (ll > r || rr < l) return;
  if (l <= ll && rr <= r)
    return tree[tr].tag = val, tree[tr].sum = (rr - ll + 1) * val, void();
  push_down(tr);
  change(l, r, tr * 2, val), change(l, r, tr * 2 + 1, val);
  push_up(tr);
}
void query_all(const int tr, const int bel) {
  int ll = tree[tr].l, rr = tree[tr].r;
  if (ll == rr) return ans[bel][ll] = tree[tr].sum, void();
  push_down(tr);
  query_all(tr * 2, bel), query_all(tr * 2 + 1, bel);
}
inline void main() {
  n = read(), m = read();
  scanf("%s", c + 1);
  for (int i = 1; i <= n; ++i) a[i] = c[i] - 'a' + 1;
  for (int i = 1; i <= m; ++i) {
    int ui = read(), vi = read(), ti = read();
    opt[i] = Node(ui, vi, ti);
  }
  for (int i = 1; i <= 26; ++i) {
    lim = i;
    build(1, n, 1);
    for (int j = 1; j <= m; ++j) {
      int cnt = query(opt[j].l, opt[j].r, 1), sum = opt[j].r - opt[j].l + 1;
      if (opt[j].type == 0)
        change(opt[j].l, opt[j].l + cnt - 1, 1, 1),
            change(opt[j].l + cnt, opt[j].r, 1, 0);
      else
        change(opt[j].l, opt[j].l + sum - cnt - 1, 1, 0),
            change(opt[j].l + sum - cnt, opt[j].r, 1, 1);
    }
    query_all(1, i);
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= 26; ++j)
      if (ans[j][i] && !ans[j + 1][i]) {
        putchar('a' - 1 + j);
        break;
      }
}
}  // namespace thecold
int main() {
  thecold::main();
  return 0;
}