#include<iostream>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
    const int  n = 200050;
    using namespace std;
    int tre[n * 4];
    void build(int num, int le, int ri)//建立线段树
    {
        if (le == ri)//如果区间左端点等于区间右端点，则输入区间的每一个元素
        {
            scanf("%d", &tre[num]);
            return;
        }
        int mid = (le + ri) / 2;
        build(num * 2, le, mid);
        build(num * 2 + 1, mid + 1, ri);
        tre[num] = max(tre[num * 2], tre[num * 2 + 1]);//通过子节点来初始化父亲节点
    }
    void update(int num, int le, int ri, int x, int y)//单点更新
    {
        if (le == ri)//找到要更新的x节点所在位置，并对x节点值加y
        {
            tre[num] = y;
            return;
        }
        int mid = (le + ri) / 2;
        if (x <= mid)
            update(num * 2, le, mid, x, y);
        else
            update(num * 2 + 1, mid + 1, ri, x, y);
        tre[num] = max(tre[num * 2], tre[num * 2 + 1]);
    }
    int query(int num, int le, int ri, int x, int y)//区间查询
    {
        if (x <= le && y >= ri)//如果我们要查询的[x,y]区间比[le,ri]大，那么[le,ri]区间的值就是结果的一部分，直接返回
        {
            return tre[num];
        }
        int mid = (le + ri) / 2;
        int ans = 0;//ans是记录查询区间的最大值
        if (x <= mid)//查询左子树
            ans = max(ans, query(num * 2, le, mid, x, y));
        if (y > mid)//查询右子树
            ans = max(ans, query(num * 2 + 1, mid + 1, ri, x, y));
        return ans;
    }
    int main()
    {
        int t, n;
        while (cin >> n >> t)
        {
            build(1, 1, n);
            char operate[10];
            for (int i = 0; i < t; i++)
            {
                cin >> operate;
                int x, y;
                scanf("%d %d", &x, &y);
                if (operate[0] == ‘U‘)
                {
                    update(1, 1, n, x, y);
                }
                if (operate[0] == ‘Q‘)
                {
                    cout << query(1, 1, n, x, y) << endl;
                }
            }
        }

        return 0;
    }