#pragma GCC optimization ("O3") #include using namespace std; using ll = long long; using vec = vector; using mat = vector; using pll = pair; #define INF (1LL<<61) #define MOD 1000000007LL // #define MOD 998244353LL #define EPS (1e-10) #define PR(x) cout << (x) << endl #define PS(x) cout << (x) << " " #define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i)) #define FORE(i,v) for(auto (i):v) #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((ll)(x).size()) #define REV(x) reverse(ALL((x))) #define ASC(x) sort(ALL((x))) #define DESC(x) {ASC((x)); REV((x));} #define BIT(s,i) (((s)>>(i))&1) #define pb push_back #define fi first #define se second template inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;} template inline int chmax(T& a, T b) {if(a=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 b(*this); return b+=a;} mint operator-(const mint& a) const {mint b(*this); return b-=a;} mint operator*(const mint& a) const {mint b(*this); return b*=a;} mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;} mint inv() const {return pow(MOD-2);} mint& operator/=(const mint& a) {return *this*=a.inv();} mint operator/(const mint& a) const {mint b(*this); return b/=a;} }; istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;} ostream &operator<<(ostream& os, const mint& a) {return os< struct SegmentTree { using FX = function; int n; FX fx; const X ex; vector dat; SegmentTree(int n_, FX fx_, X ex_) : n(), fx(fx_), ex(ex_) { int x = 1; while(n_ > x) x *= 2; n = x; dat = vector(n*2, ex); } void set(int i, X x) { dat[i+n-1] = x; } void build() { for (int k = n-2; k>=0; k--) dat[k] = fx(dat[k*2+1], dat[k*2+2]); } void update(int i, X x) { i += n-1; dat[i] = x; while(i > 0) { i = (i-1)/2; dat[i] = fx(dat[i*2+1], dat[i*2+2]); } } X output_sub(int a, int b, int k, int l, int r) { if(r <= a || b <= l) return ex; else if(a <= l && r <= b) return dat[k]; else { X vl = output_sub(a, b, k*2+1, l, (l+r)/2); X vr = output_sub(a, b, k*2+2, (l+r)/2, r); return fx(vl, vr); } } X output(int a, int b) { return output_sub(a, b, 0, 0, n); } int find_rightest_sub(int a, int b, X x, int k, int l, int r) { if(dat[k] < x || r <= a || b <= l) return a-1; else if(k >= n-1) return k-(n-1); else { int vr = find_rightest_sub(a, b, x, 2*k+2, (l+r)/2, r); if(vr != a-1) return vr; else return find_rightest_sub(a, b, x, 2*k+1, l, (l+r)/2); } } int find_rightest(int a, int b, X x) { return find_rightest_sub(a, b, x, 0, 0, n); } int find_leftest_sub(int a, int b, X x, int k, int l, int r) { if(dat[k] < x || r <= a || b <= l) return b; else if(k >= n-1) return k-(n-1); else { int vl = find_leftest_sub(a, b, x, 2*k+1, l, (l+r)/2); if(vl != b) return vl; else return find_leftest_sub(a, b, x, 2*k+2, (l+r)/2, r); } } int find_leftest(int a, int b, X x) { return find_leftest_sub(a, b, x, 0, 0, n); } }; int main() { ll N, Q; cin >> N >> Q; ll ex = INF; auto fx = [](ll l, ll r) {return min(l, r);}; SegmentTree st(N+1, fx, ex); vec I(N+1); REP(i,1,N+1) { ll a; cin >> a; st.set(i, a); I[a] = i; } st.build(); while(Q--) { ll t, l, r; cin >> t >> l >> r; if(t == 1) { ll xl = st.output(l, l+1); ll xr = st.output(r, r+1); st.update(l, xr); st.update(r, xl); swap(I[xl], I[xr]); } else { ll m = st.output(l, r+1); PR(I[m]); } } return 0; } /* */