#include<stdio.h>
#include<vector>
using namespace std;

const int NMAX = (1<<20) + 5, MOD = 666013;

int n, distinct, x[NMAX];
vector < pair <int, int> > hash[MOD];
vector < pair <int, int> >::iterator it;

void del (int val) {
    int ln = val % MOD;

    for(it = hash[ln].begin(); it != hash[ln].end(); ++ it)
        if(it->first == val) {
            -- it->second;
            if(it->second == 0)
                -- distinct;
        }
}

void ins (int val) {
    int ln = val % MOD, found = 0;
    for(it = hash[ln].begin(); it != hash[ln].end(); ++ it)
        if(it->first == val) {
            ++ it->second;
            if(it->second == 1)
                ++ distinct;
            found = 1;
        }
    if(!found)
        hash[ln].push_back(make_pair(val, 1)),
        ++ distinct;
}

int cnt (int lim) {
    int i, left, right, ans;
    
    distinct = 0;
    for(i = 0; i < MOD; ++ i)
        hash[i].clear();

    left = 1;
    right = ans = 0;
    while(left <= n && right <= n ) {
        while(distinct < lim && right < n) {
            ins(x[right + 1]);
            ++ right;
        }
        ans += right - left + 1;
        del(x[left]);
        ++ left;
    }
    return ans;
}

int main() {
    freopen("secv5.in", "r", stdin);
    freopen("secv5.out", "w", stdout);
    int l, u, i;

    scanf("%d%d%d", &n, &l, &u);
    for(i = 1; i <= n; ++ i)
        scanf("%d", &x[i]);

    printf("%d\n", cnt(u) - cnt(l - 1));
    return 0;
}