#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;
const int N = 100010;

#define ls u << 1
#define rs u << 1 | 1

struct Node {
    int l, r;
    int add, sum;
} tr[N << 2];

void pushup(int u) {
    tr[u].sum = tr[ls].sum | tr[rs].sum; //因为是模拟二进制的操作，所以采用 或 运算即可完成拼接操作
}

void pushdown(int u) {
    if (tr[u].add) {
        tr[ls].add = tr[rs].add = tr[u].add;
        tr[ls].sum = tr[rs].sum = tr[u].add;
        tr[u].add = 0;
    }
}

void build(int u, int l, int r) {
    tr[u].l = l, tr[u].r = r;
    if (l == r) {
        tr[u].sum = 1; // sum记录的其实是一个30位的二进制压缩模拟数字，此时置为1表示最后一位即0位为1，是默认的颜色
        return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
    pushup(u);
}

void modify(int u, int l, int r, int c) {
    if (l <= tr[u].l && tr[u].r <= r) {
        tr[u].add = 1 << (c - 1);
        tr[u].sum = 1 << (c - 1);
        return;
    }

    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) modify(u << 1, l, r, c);
    if (r > mid) modify(u << 1 | 1, l, r, c);
    pushup(u);
}

int query(int u, int l, int r) {
    if (l <= tr[u].l && tr[u].r <= r) return tr[u].sum;

    int ans = 0;
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) ans = query(u << 1, l, r);
    if (r > mid) ans = ans | query(u << 1 | 1, l, r);
    return ans;
}

//手工版本数字1个数
int count(int x) {
    int cnt = 0;
    while (x) {
        cnt++;
        x -= (x & -x);
    }
    return cnt;
}

int main() {
    //加快读入
    ios::sync_with_stdio(false), cin.tie(0);
    int n, t, m; //区间[1,n],t种颜色,m个操作
    cin >> n >> t >> m;

    //构建
    build(1, 1, n);

    while (m--) {
        char op;
        int l, r;
        cin >> op >> l >> r;
        if (l > r) swap(l, r); //坑点题目没说哪个大哪个小，需要特判
        if (op == 'C') {
            int c;
            cin >> c;
            modify(1, l, r, c);
        } else {
            int ans = query(1, l, r);
            int cnt = count(ans); //获取某个数二进制位1的个数
            printf("%d\n", cnt);
        }
    }
    return 0;
}