#include <bits/stdc++.h> // By Lucky Ox
#define int long long
#define endl "\n"
#define pii pair<int, int>
#define PI atan(1.0) * 4
using namespace std;
using i128 = __int128;
typedef unsigned long long ull;
const int mod = LONG_LONG_MAX, INF = 0x3f3f3f3f3f3f3f3f;
int P(int x, int p){ return (x % p + p) % p; }
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len; }
int q_pow(int a, int k, int p) { int res = 1; while (k) { if (k & 1) res = res * a % p; k >>= 1; a = a * a % p; } return res; }
int to_int(string s) { int val = 0; for(int i = 0; i < (int)s.size(); i ++ ){val *= 10; val += s[i] - '0';}return val;}//注意:s是空串也会返回0
i128 read() { i128 x = 0; char c = getchar(); while (c < '0' || c > '9') c = getchar(); while (c >= '0' && c <= '9') {
x = x * 10 + c - '0'; c = getchar(); } return x; } //i128输入
void print(i128 x) {if(x > 9) print(x / 10); putchar(x % 10 + '0'); }//i128输出
//用__lg()来求一个数二进制下的位数 返回的len 表示这个数是[0, 1, ...., len] 比如10 __lg(10) = 3, 1010 [3, 2, 1, 0]
//__builtin_popcountll(int x) 求二进制下x中1的数量 __buitlin_ctzll(int x) 求二进制下末尾0的个数
//(n & (1 << i))的值可能会是1, 2, 4...... (n >> i) & 1的值一定是1
//将x转换成[1, n]中对应的数 (x - 1) % n + 1
//bool __ED__;
const int N = 1e2 + 10;
int n, l, r;
int a[N], s[N];
int dp[N][N][N]; // 元素数量为k 区间i，j的最小花费
signed main() {
    //cerr << abs(&__ED__-&__ST__) / (1024.0 * 1024.0) <<" MiB\n";
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(10);
    while(cin >> n >> l >> r) {
        for(int i = 1; i <= n; i ++ ) cin >> a[i], s[i] = s[i - 1] + a[i];

        memset(dp, 0x3f, sizeof dp);
        for(int i = 1; i <= n; i ++ ) dp[1][i][i] = 0;

        for(int len = 2; len <= n; len ++ ) {
            for(int i = 1; i + len - 1 <= n; i ++ ) {
                int j = i + len - 1;
                for(int k = 1; k <= len; k ++ ) {
                    if(k == 1) {
                        for(int cj = l; cj <= r; cj ++ ) {
                            dp[k][i][j] = min(dp[k][i][j], dp[cj][i][j] + s[j] - s[i - 1]);
                        }
                    }
                    else {
                        for(int cj = i; cj <= j - 1; cj ++ ) {
                            dp[k][i][j] = min(dp[k][i][j], dp[1][i][cj] + dp[k - 1][cj + 1][j]);
                        }
                    }
                }
            }
        }
        if(dp[1][1][n] >= 0x3f3f3f3f3f) cout << 0 << endl;
        else cout << dp[1][1][n] << endl;
    }
    return 0;
}
/*
3 2 2
1 2 3
3 2 3
1 2 3
4 3 3 
1 2 3 4
*/