The unlucky Ikta-kun has rewritten the important character string T that he had to a different character string T'by the virus. It is known that the virus has rewritten one letter of T to a different letter. That is, T and T'are different by exactly one character. In order to restore T, Ikta prepared a document S in which T appears. As a preparation for restoring T, I would like to find out the number of substrings of S that may match T.

Given the string T'and the document S. S = a1 a2 a3 .. .a | S | length | T'| substring ak ak + 1 ... ak + | T'| −1 (1 ≤ k ≤ | S | − | T'| Find the number of things that differ by one character from T'in +1).



Input

The input is given in the following format.


S
T'


* S is given in the first line.
* T'is given in the second line.
* S and T'consist only of uppercase and lowercase alphabets, respectively.



Constraints

Each variable being input satisfies the following constraints.

* 1 ≤ | S | ≤ 300,000
* 1 ≤ | T'| ≤ | S |

Output

Output the number of substrings that satisfy the conditions on one line.

Examples

Input

abcbcdbc
abc


Output

2


Input

aaaaaa
aaaaaa


Output

0


Input

baaaaaaaa
b


Output

8
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (1e+18) + 7;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

ll mod_pow(ll a, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%mod;
		a = a * a%mod; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

//const int max_n = 1 << 22;
//modint fact[max_n], factinv[max_n];
//void init_f() {
//	fact[0] = modint(1);
//	for (int i = 0; i < max_n - 1; i++) {
//		fact[i + 1] = fact[i] * modint(i + 1);
//	}
//	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
//	for (int i = max_n - 2; i >= 0; i--) {
//		factinv[i] = factinv[i + 1] * modint(i + 1);
//	}
//}
//modint comb(int a, int b) {
//	if (a < 0 || b < 0 || a < b)return 0;
//	return fact[a] * factinv[b] * factinv[a - b];
//}
void Z_algorithm(const string &s, vector<int> &a) {
	int sz = s.size();
	a.resize(sz);
	a[0] = sz;
	int i = 1, j = 0;
	while (i < sz) {
		while (i + j < sz&&s[j] == s[i + j])++j;
		a[i] = j;
		if (j == 0) { ++i; continue; }
		int k = 1;
		while (i + k < sz&&k + a[k] < j)a[i + k] = a[k], ++k;
		i += k; j -= k;
	}
}
void solve() {
	string s, t; cin >> s >> t;
	int n = s.size();
	int m = t.size();
	vector<int> ri(n),le(n);
	string u = t + s;
	vector<int> ra;
	Z_algorithm(u, ra);
	rep(i, n)ri[i] = ra[i + t.size()];
	u = s + t;
	reverse(all(u));
	ra.clear();
	Z_algorithm(u, ra);
	rep(i, n) {
		le[i] = ra[n + t.size() - 1 - i];
	}
	int ans = 0;
	rep(i, n - m + 1) {
		int sum = ri[i] + le[i + m - 1];
		if (sum == m - 1)ans++;
	}
	cout << ans << endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(12);
	//init_f();
	//int t; cin >> t; rep(i, t)solve();
	solve();
	stop
		return 0;
}

