#include <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 998244353;
const ll MAX = 4000001;
const double eps = 1E-18;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint operator^(const ll& t)const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}
	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

///////////////////////////
template<typename T>
struct Segment_Tree {
	typedef function<T(T, T)> F;

	ll n;
	F f;//モノイドとしての演算(最小値、和　など)
	T unit;//単位元

	ll size;
	vector<T> seg;

	Segment_Tree(ll n0, F f0, T unit0) {
		n = n0;
		f = f0;
		unit = unit0;

		size = 1;
		while (size < n) {
			size *= 2;
		}
		seg.assign(2 * size, unit);
	}


	void update(ll k, T x) {//k番目の要素をxに更新
		k += size - 1;
		seg[k] = x;
		while (k > 1) {
			k /= 2;
			seg[k] = f(seg[2 * k], seg[2 * k + 1]);
		}
	}



	T query_prep(ll a, ll b, ll k, ll l, ll r) {//[a,b)とkに対応する区間[l,r)の共通部分でクエリ
		if (b <= l || a >= r) { return unit; }
		if (a <= l && r <= b) { return seg[k]; }

		T v1 = query_prep(a, b, 2 * k, l, (l + r) / 2);
		T v2 = query_prep(a, b, 2 * k + 1, (l + r) / 2, r);
		return f(v1, v2);

	}

	T query(ll a, ll b) {
		return query_prep(a, b, 1, 1, size + 1);
	}

};




int main() {
	string S;
	cin >> S;
	ll N = S.size();
	ll K;
	cin >> K;

	vector<vector<vector<ll>>> dp(N + 1, vector<vector<ll>>(N + 1, vector<ll>(K + 1, 0)));

	repn(i, N)repn(j, N)rep(k, K + 1) {
		if (i + j > N) { continue; }

		if (S[i-1] == S[N-j]) {
			dp[i][j][k] = max(dp[i - 1][j][k], max(dp[i][j - 1][k], dp[i - 1][j - 1][k] + 1));
		}
		else {
			dp[i][j][k] = max(dp[i - 1][j][k], dp[i][j - 1][k]);
			if (k > 0) { dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j - 1][k - 1] + 1); }
		}
	}

	ll ans = 0;
	rep(i, N + 1) {
		ans = max(ans, dp[i][N - i][K]*2);
	}
	rep(i, N) {
		ans = max(ans, dp[i][N - 1 - i][K] * 2 + 1);
	}
	cout << ans << endl;
	
	
	system("PAUSE");
}


Measured memory_bytes: 217076