Problem:
Range Minimum Query (RMQ)

Write a program which manipulates a sequence A = {a0, a1, . . . , an-1} with the following operations:

find(s, t): report the minimum element in as, as+1, . . . ,at.

update(i, x): change ai to x.

Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Input

n q
com0 x0 y0
com1 x1 y1
...
comq−1 xq−1 yq−1

In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes update(xi, yi) and '1' denotes find(xi, yi).

Output

For each find operation, print the minimum element.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 100000

If comi is 0, then 0 ≤ xi < n, 0 ≤ yi < 231-1.

If comi is 1, then 0 ≤ xi < n, 0 ≤ yi < n.

Sample Input 1

3 5
0 0 1
0 1 2
0 2 3
1 0 2
1 1 2

Sample Output 1

1
2

Sample Input 2

1 3
1 0 0
0 0 5
1 0 0

Sample Output 2

2147483647
5

Solution in C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

struct Node {
  int v;
  Node(int v) : v(v) {}

  Node operator*(const Node &r) const {
    return Node(min(v, r.v));
  }
} e((1ll << 31) - 1);

struct SegmentTree {
  size_t n;
  vector<Node> v;

  SegmentTree(size_t size) : n(1 << (32 - __builtin_clz(size - 1))), v(2 * n - 1, e) {}

  void update(int k, const Node &x) {
    k += n - 1;
    v[k] = x;
    while (k) {
      k = (k - 1) / 2;
      v[k] = v[2 * k + 1] * v[2 * k + 2];
    }
  }

  Node query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return e;
    if (a <= l && r <= b) return v[k];
    return query(a, b, 2 * k + 1, l, (l + r) / 2) * query(a, b, 2 * k + 2, (l + r) / 2, r);
  }

  void update(int k, int x) {
    update(k, Node(x));
  }

  int query(int a, int b) {
    return query(a, b, 0, 0, n).v;
  }
};

int main() {
  int n, q;
  cin >> n >> q;

  SegmentTree seg(n);
  while (q--) {
    int c, x, y;
    cin >> c >> x >> y;
    if (c)
      cout << seg.query(x, y + 1) << endl;
    else
      seg.update(x, y);
  }
}