Problem: There are n cities along the road, which can be represented as a straight line. The i-th city is situated at the distance of a_i kilometers from the origin. All cities are situated in the same direction from the origin. There are m trucks travelling from one city to another. 

Each truck can be described by 4 integers: starting city s_i, finishing city f_i, fuel consumption c_i and number of possible refuelings r_i. The i-th truck will spend c_i litres of fuel per one kilometer. 

When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The i-th truck can be refueled at most r_i times. Each refueling makes truck's gas-tank full. All trucks start with full gas-tank.

All trucks will have gas-tanks of the same size V litres. You should find minimum possible V such that all trucks can reach their destinations without refueling more times than allowed.

Input

First line contains two integers n and m (2 ≤ n ≤ 400, 1 ≤ m ≤ 250000) — the number of cities and trucks.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i < a_{i+1}) — positions of cities in the ascending order.

Next m lines contains 4 integers each. The i-th line contains integers s_i, f_i, c_i, r_i (1 ≤ s_i < f_i ≤ n, 1 ≤ c_i ≤ 10^9, 0 ≤ r_i ≤ n) — the description of the i-th truck.

Output

Print the only integer — minimum possible size of gas-tanks V such that all trucks can reach their destinations.

Example

Input


7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2


Output


55

Note

Let's look at queries in details: 

  1. the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50. 
  2. the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48. 
  3. the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52. 
  4. the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it's optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40. 
  5. the 5-th truck has the same description, so it also needs gas-tank of volume at least 40. 
  6. the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55. 

Solution (in JAVA):

Solution:
/**
 * author: derrick20
 * created: 11/26/20 10:37 PM
 */
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class TrucksAndCitiesFast {
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        
        int N = sc.nextInt();
        int M = sc.nextInt();
        long[] a = sc.nextLongs(N);
        long limitingCapacity = 0;
        ArrayList<int[]>[] queries = new ArrayList[N + 1];
        Arrays.setAll(queries, i -> new ArrayList<>());
        for (int i = 0; i < M; i++) {
            int start = sc.nextInt() - 1;
            int finish = sc.nextInt() - 1;
            int useRate = sc.nextInt();
            int refuels = sc.nextInt();
            queries[refuels].add(new int[]{start, finish, useRate});
        }
        long[][][] dp = new long[2][N][N];
        int curr = 0;
        for (int k = 0; k <= N; k++) {
//            System.out.println("For k = " + k);
            int next = 1 ^ curr;
            for (int l = 0; l <= N - 1; l++) {
                /*
                Key idea:
                Now, because the optimal structure of the partition is such
                that the last rest stop is non-decreasing, we can solve n values
                of r in O(n) time total, rather than O(n) for each.

                Our easy transition is:
                min over all last vals of {max(dp[l][last][k - 1], a[r] - a[last]}
                Now, we just need to keep moving forward while that's non-increasing

                Need to go in order of increasing k
                 */
                dp[next][l][l] = 0;
                int last = l;
                for (int r = l + 1; r <= N - 1; r++) {
                    if (k == 0) {
                        dp[next][l][r] = a[r] - a[l];
                    } else {
                        while (last + 1 <= r && max(dp[curr][l][last], a[r] - a[last]) >= max(dp[curr][l][last + 1], a[r] - a[last + 1])) {
                            last++;
                        }
                        dp[next][l][r] = max(dp[curr][l][last], a[r] - a[last]);
                    }
//                    System.out.println(l + " " + r + " => " + dp[l][r][k]);
                }
            }
            curr = next;
            for (int[] query : queries[k]) {
                // Very nice problem
                int start = query[0];
                int finish = query[1];
                long useRate = query[2];
                long requiredCapacity = useRate * dp[curr][start][finish];
                limitingCapacity = max(limitingCapacity, requiredCapacity);
            }
        }
        out.println(limitingCapacity);
        out.close();
    }

    static class FastScanner {
        private int BS = 1 << 16;
        private char NC = (char) 0;
        private byte[] buf = new byte[BS];
        private int bId = 0, size = 0;
        private char c = NC;
        private double cnt = 1;
        private BufferedInputStream in;

        public FastScanner() {
            in = new BufferedInputStream(System.in, BS);
        }

        public FastScanner(String s) {
            try {
                in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
            } catch (Exception e) {
                in = new BufferedInputStream(System.in, BS);
            }
        }

        private char getChar() {
            while (bId == size) {
                try {
                    size = in.read(buf);
                } catch (Exception e) {
                    return NC;
                }
                if (size == -1) return NC;
                bId = 0;
            }
            return (char) buf[bId++];
        }

        public int nextInt() {
            return (int) nextLong();
        }

        public int[] nextInts(int N) {
            int[] res = new int[N];
            for (int i = 0; i < N; i++) {
                res[i] = (int) nextLong();
            }
            return res;
        }

        public long[] nextLongs(int N) {
            long[] res = new long[N];
            for (int i = 0; i < N; i++) {
                res[i] = nextLong();
            }
            return res;
        }

        public long nextLong() {
            cnt = 1;
            boolean neg = false;
            if (c == NC) c = getChar();
            for (; (c < '0' || c > '9'); c = getChar()) {
                if (c == '-') neg = true;
            }
            long res = 0;
            for (; c >= '0' && c <= '9'; c = getChar()) {
                res = (res << 3) + (res << 1) + c - '0';
                cnt *= 10;
            }
            return neg ? -res : res;
        }

        public double nextDouble() {
            double cur = nextLong();
            return c != '.' ? cur : cur + nextLong() / cnt;
        }

        public double[] nextDoubles(int N) {
            double[] res = new double[N];
            for (int i = 0; i < N; i++) {
                res[i] = nextDouble();
            }
            return res;
        }

        public String next() {
            StringBuilder res = new StringBuilder();
            while (c <= 32) c = getChar();
            while (c > 32) {
                res.append(c);
                c = getChar();
            }
            return res.toString();
        }

        public String nextLine() {
            StringBuilder res = new StringBuilder();
            while (c <= 32) c = getChar();
            while (c != '\n') {
                res.append(c);
                c = getChar();
            }
            return res.toString();
        }

        public boolean hasNext() {
            if (c > 32) return true;
            while (true) {
                c = getChar();
                if (c == NC) return false;
                else if (c > 32) return true;
            }
        }
    }
    static void ASSERT(boolean assertion, String message) {
        if (!assertion) throw new AssertionError(message);
    }
    static void ASSERT(boolean assertion) {
        if (!assertion) throw new AssertionError();
    }
}