Problem:
Score : 400 points

Problem Statement

There are N towns located in a line, conveniently numbered 1 through N. Takahashi the merchant is going on a travel from town 1 to town N, buying and selling apples.

Takahashi will begin the travel at town 1, with no apple in his possession. The actions that can be performed during the travel are as follows:

Move: When at town i (i < N), move to town i + 1.

Merchandise: Buy or sell an arbitrary number of apples at the current town. Here, it is assumed that one apple can always be bought and sold for A_i yen (the currency of Japan) at town i (1 ≦ i ≦ N), where A_i are distinct integers. Also, you can assume that he has an infinite supply of money.

For some reason, there is a constraint on merchandising apple during the travel: the sum of the number of apples bought and the number of apples sold during the whole travel, must be at most T. (Note that a single apple can be counted in both.)

During the travel, Takahashi will perform actions so that the profit of the travel is maximized. Here, the profit of the travel is the amount of money that is gained by selling apples, minus the amount of money that is spent on buying apples. Note that we are not interested in apples in his possession at the end of the travel.

Aoki, a business rival of Takahashi, wants to trouble Takahashi by manipulating the market price of apples. Prior to the beginning of Takahashi's travel, Aoki can change A_i into another arbitrary non-negative integer A_i' for any town i, any number of times. The cost of performing this operation is |A_i - A_i'|. After performing this operation, different towns may have equal values of A_i.

Aoki's objective is to decrease Takahashi's expected profit by at least 1 yen. Find the minimum total cost to achieve it. You may assume that Takahashi's expected profit is initially at least 1 yen.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9 (1 ≦ i ≦ N)

A_i are distinct.

2 ≦ T ≦ 10^9

In the initial state, Takahashi's expected profit is at least 1 yen.

Input

The input is given from Standard Input in the following format:

N T
A_1 A_2 ... A_N

Output

Print the minimum total cost to decrease Takahashi's expected profit by at least 1 yen.

Sample Input 1

3 2
100 50 200

Sample Output 1

1

In the initial state, Takahashi can achieve the maximum profit of 150 yen as follows:

Move from town 1 to town 2.

Buy one apple for 50 yen at town 2.

Move from town 2 to town 3.

Sell one apple for 200 yen at town 3.

If, for example, Aoki changes the price of an apple at town 2 from 50 yen to 51 yen, Takahashi will not be able to achieve the profit of 150 yen. The cost of performing this operation is 1, thus the answer is 1.

There are other ways to decrease Takahashi's expected profit, such as changing the price of an apple at town 3 from 200 yen to 199 yen.

Sample Input 2

5 8
50 30 40 10 20

Sample Output 2

2

Sample Input 3

10 100
7 10 4 5 9 3 6 8 2 1

Sample Output 3

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <stdbool.h>
#include <limits.h>
#include <cinttypes>
#include <string.h>
#include <bits/stdc++.h>

using namespace std;

#define countof(a) (sizeof(a)/sizeof(*a))

#define vi vector<int>
#define vvi vector<vector<int> >
#define vpi vector<pi >
#define pi pair<int,int>
#define fi first
#define se second
#define all(n) n.begin(), n.end()

#define FOR(var, to)          for (register s64 var = 0; var < (s64)to; var++)
#define FROMTO(var, from, to) for (register s64 var = from; var <= (s64)to; var++)

#define INIT(var) FOR(i,countof(var)) var[i] = 0
#define INPUT(var) FOR(i,countof(var)) var[i] = ri()

#define SORT(v) qsort(v,countof(v),sizeof(*v),int_less)
#define SORTT(v) qsort(v,countof(v),sizeof(*v),int_greater)
#define QSORT(v,b) qsort(v,countof(v),sizeof(*v),b)

#define MOD 1000000007
#define INF (1 << 30)

typedef uint8_t u8;
typedef uint16_t u16;
typedef uint32_t u32;
typedef uint64_t u64;
typedef int8_t s8;
typedef int16_t s16;
typedef int32_t s32;
typedef int64_t s64;

template<int mod>
struct ModInt{
    int x;
    ModInt():x(0){}
    ModInt(long long y):x(y>=0?y%mod:(mod-(-y)%mod)%mod){}
    ModInt &operator+=(const ModInt &p){
        if((x+=p.x)>=mod)x-=mod;
        return *this;
    }
    ModInt &operator-=(const ModInt &p){
        if((x+=mod-p.x)>=mod)x-=mod;
        return *this;
    }
    ModInt &operator*=(const ModInt &p){
        x=(int)(1LL*x*p.x%mod);
        return *this;
    }
    ModInt &operator/=(const ModInt &p){
        *this*=p.inverse();
        return *this;
    }
    ModInt operator-()const{return ModInt(-x);}
    ModInt operator+(const ModInt &p)const{return ModInt(*this)+=p;}
    ModInt operator-(const ModInt &p)const{return ModInt(*this)-=p;}
    ModInt operator*(const ModInt &p)const{return ModInt(*this)*=p;}
    ModInt operator/(const ModInt &p)const{return ModInt(*this)/=p;}
    bool operator==(const ModInt &p)const{return x==p.x;}
    bool operator!=(const ModInt &p)const{return x!=p.x;}
    ModInt inverse()const{
        int a=x,b=mod,u=1,v=0,t;
        while(b>0){
            t=a/b;
            a-=t*b;
            swap(a,b);
            u-=t*v;
            swap(u,v);
        }
        return ModInt(u);
    }
    friend ostream &operator<<(ostream &os,const ModInt<mod> &p){
        return os<<p.x;
    }
    friend istream &operator>>(istream &is,ModInt<mod> &a){
        long long x;
        is>>x;
        a=ModInt<mod>(x);
        return (is);
    }
};
typedef ModInt<MOD> mint;

struct UnionFind{
    vi data;
    UnionFind(int size):data(size,-1){}
    bool unite(int x,int y) {
        x=root(x);y=root(y);
        if(x!=y){
            if(data[y]<data[x])swap(x,y);
            data[x]+=data[y];data[y]=x;
        }
        return x!=y;
    }
    bool find(int x,int y) {
        return root(x)==root(y);
    }
    int root(int x) {
        return data[x]<0?x:data[x]=root(data[x]);
    }
    int size(int x) {
        return -data[root(x)];
    }
    bool united() {
        int comroot = -1;
        FOR(i,data.size()) {
            if (comroot != -1 && root(i) != comroot) return false;
            comroot = root(i);
        }
        return true;
    }
};

static inline int ri() {
  int a;
  scanf("%d", &a);
  return a;
}

static inline s64 rs64() {
  s64 a;
  scanf("%" SCNd64, &a);
  return a;
}

static inline void wi(int a) {
  printf("%d", a);
}

static inline void wu64(u64 a) {
  printf("%" PRIu64, a);
}

static inline void ws64(s64 a) {
  printf("%" PRId64, a);
}

int int_less(const void *a, const void *b) {
  return (*((const int*)a) - *((const int*)b));
}
int int_greater(const void *a, const void *b) {
  return (*((const int*)b) - *((const int*)a));
}

int main() {
    int n,t;
    cin >> n >> t;
    int a[n];
    INPUT(a);
    int res = 0;
    int dis_max = -1;
    int dismaxi = -1;
    int m0 = -1;
    int m0i = -1;
    int m1 = -1;
    FOR(i,n) {
        if (m0 == -1) m0 = a[i], m0i = i;
        if (m1 == -1) m1 = a[i];
        if (a[i] <= m0) m0 = a[i], m0i = i, m1 = a[i];
        if (a[i] >= m1) m1 = a[i];
        int dis_new = m1-m0;
        if (dis_max == dis_new) {
            if (dismaxi != m0i) res++;
        }
        if (dis_max < dis_new) dis_max = dis_new, dismaxi = i, res = 1;
    }
        cout << res << endl;
    
  return 0;
}